Was he right?

[Phil]

An up-and-comer faced a tricky play problem in 5♣:

 

[hv=pc=n&s=sJ2h2dqt42caj9852&n=sq63hak43dakj3cq3&d=e&v=0&b=14&a=1sp2sd3s(Competitive)5cppp]266|200[/hv]

 

The opening lead was the ♠2 won by the Ace. A spade was returned to the K, and the defense exited with the ♦7.

 

Declarer led the ♣Q covered by the King and won. Declarer returned to dummy with a heart and played a 2nd club. His luck was in when he rose with the J and dropped the ♣T offside doubleton.

 

Wow!, said his partner.

 

"Nothing to it" said the UAC. I played for this layout because my LHO had more vacant spaces.

 

Was he right?

[ggwhiz]

Maybe he's just tall? My pard once declared when his rho told their pard to hold their hand up. He said "too late".

 

However when his lho followed to the first trump, the size of the spot matters. You can't pick up KTxx onside and few defenders will falsecard with the 7 from 7x.

[ahydra]

That auction cannot be right. Did South bid 5C over 3S?

 

And I'm confused about the vacant spaces thing as well. W has more vacant spaces, yes, but when E follows to the 2nd round then he knows that trumps are 3-2, 2-3 or 1-4. Is he arguing "if they're 3-2 I'm screwed anyway; of the remainder, vacant spaces suggests they are 2-3 rather than 1-4; assuming a 2-3 split is the case, W has 10x"? I can follow that all the way up until the last step, since with a 2-3 split East is surely the favourite to hold the ten.

 

ahydra

[Phil]

Maybe he's just tall? My pard once declared when his rho told their pard to hold their hand up. He said "too late".

 

However when his lho followed to the first trump, the size of the spot matters. You can't pick up KTxx onside and few defenders (his lho) will falsecard with the 7 from T7x.

 

These opponents are good enough to randomize their spots, and LHO is DEFINITELY good enough to play his highest spot when he's not holding the T.

[Phil]

That auction cannot be right. Did South bid 5C over 3S?

 

 

Thanks, fixed.

[aguahombre]

Dunno if the play was correct in theory, but he certainly was right.

[lalldonn]

Um, he was not right lol. His play was "right" because it worked, but his reasoning was wrong!

[Phil]

Um, he was not right lol. His play was "right" because it worked, but his reasoning was wrong!

 

Then please tell the world your reasoning and 'why' he was wrong!

[ggwhiz]

These opponents are good enough to randomize their spots, and LHO is DEFINITELY good enough to play his highest spot when he's not holding the T.

 

But did they play the 7, restricted choice? Otherwise, right guess wrong reasons.

[lalldonn]

Then please tell the world your reasoning and 'why' he was wrong!

His play only matters if clubs are exactly 2-3. "Vacant spaces" is an argument to determine the likelihood of a particular break (like, LHO has more vacant spaces so he is probably longer in clubs than RHO), not what to do when you already know you are getting that break.

 

There is also no restricted choice argument. They have both played one spot so the argument applies the same to both opponents and cancels out.

 

The clubs other than the king are 2-2 meaning only clues you have to go on are psychological (outguessing the opponents on which spot they played) or if you think a cover is more/less likely from Kxx than KTx. The only technical reason is a potential trump coup (RHO 5134 and LHO 3721?) which they have defended well by not playing a third spade so it won't work.

[y66]

We can not use vacant spaces to calculate the odds of RHO holding KTx vs Kxx, the only relevant holdings after RHO plays a second club.

[silvr bull]

It is ALWAYS right to be lucky! :D

I think his choice of words was unfortunate, but it is right to play for the drop. After the ♣QKAx fall on the first ♣ trick, the situation is exactly the same as if the original ♣ holding was as shown below:

♣Kxx

♣AJxxxx

 

After playing the ♣K and both follow, do you try to drop the ♣Q or finesse? The odds slightly favor the drop, unless LHO has shown length in a side suit. In this case, RHO has shown the side suit length, so the odds are even higher for the drop.

[rmnka447]

Vacant places only strictly applies when using the suit or suits for which the distribution is COMPLETELY known to determine the vacant places remaining in the unseen hands. From the bidding, play and Declarer/Dummy holdings, the only suit you know completely at the moment of decision in Clubs is Spades. East started with 5 and West with 3. So, that leaves East with 8 vacant places and West with 10 vacant places. So the odds are 10 to 8 (56% probabilty) that West holds any particular card -- in this case -- the Club 10.

 

So it seems as if UAC has made the best play to make the contract.

[655321]

 

So the odds are 10 to 8 (56% probabilty) that West holds any particular card -- in this case -- the Club 10.

 

So it seems as if UAC has made the best play to make the contract.

 

No, your 56% includes the times West has 10xx as well - read the lalldonn post above.

[Fluffy]

No need to be right when you can bid 4NT and let partner play 5♦. Also worth thinking about would be double followed by 5♣, but I don't know if it should mean something like this.

[lowerline]

An up-and-comer faced a tricky play problem in 5♣:

 

[hv=pc=n&s=sJ2h2dqt42caj9852&n=sq63hak43dakj3cq3&d=e&v=0&b=14&a=1sp2sd3s(Competitive)5cppp]266|200[/hv]

 

The opening lead was the ♠2 won by the Ace. A spade was returned to the K, and the defense exited with the ♦7.

 

Declarer led the ♣Q covered by the King and won. Declarer returned to dummy with a heart and played a 2nd club. His luck was in when he rose with the J and dropped the ♣T offside doubleton.

 

Wow!, said his partner.

 

"Nothing to it" said the UAC. I played for this layout because my LHO had more vacant spaces.

 

Was he right?

 

So East opened with 10hcp and West raised with 3hcp? Then I suspect they don't have flat distributions...

Based on the knowledge that East started with 5 spades and West with 3, declarer decided that *EAST* is more likely to hold 3 clubs without the ten? I don't get it...

 

Steven

[phil_20686]

The only relevant holdings are Tx Kxx, xx KTx, there are three holdings of each, so its exactly 50%.

 

If you are unsure of how to do these problems, looking at the a priori percentages of all the holdings that are still relevant is the certain way of getting the right answer.

[phil_20686]

Vacant places only strictly applies when using the suit or suits for which the distribution is COMPLETELY known to determine the vacant places remaining in the unseen hands. From the bidding, play and Declarer/Dummy holdings, the only suit you know completely at the moment of decision in Clubs is Spades. East started with 5 and West with 3. So, that leaves East with 8 vacant places and West with 10 vacant places. So the odds are 10 to 8 (56% probabilty) that West holds any particular card -- in this case -- the Club 10.

 

So it seems as if UAC has made the best play to make the contract.

 

This reasoning is fallacious. The vacant spaces arguments applies only to suits whose break is not known. The reason lho is more likely to hold a specific card in a suit is precisely because he is more likely to hold more cards in that suit.

 

I agree that lho is more likely to hold the club ten, but that is precisely because the breaks that are still possible include Txx-Kx, however, that break cannot be picked up, so it is irrelevant.

[Phil]

Of course he was wrong. He made a lucky guess.

 

For a better explanation, check out "Expert Bridge Simplified", by Jeff Rubens.

 

Rubens calls the non-K clubs a 'suit within a suit' or a 'subsuit'. There are four non-king clubs, and they will split 2-2 more often than 3-1. We can't cope with a 3-1 remainder either way, so it more or less mirrors a two way finesse.

 

For those that like to crunch numbers, you can figure out the relative chances of LHO having three clubs versus two based on the known vacant spaces.

[JLOGIC]

A simpler way of looking at it is that there are 3 clubs remaining, and your play only matters if LHO has a singleton remaining. 2 of the clubs are small and 1 is the ten, so it is 2:1 to hook.

 

More advanced players will get hung up on restricted choice elements. Yes, there is restricted choice on LHO to begin with because if he had 2 small he could play either, but with Tx he would have to play the x. However, there is also the same restricted choice on RHO now, if he had 2 small clubs remaining he could play either, but if he had Tx he would have to play the x, so those cancel out.

[phil_20686]

A simpler way of looking at it is that there are 3 clubs remaining, and your play only matters if LHO has a singleton remaining. 2 of the clubs are small and 1 is the ten, so it is 2:1 to hook.

 

This logic feels intuitive, but is actually wrong. Suppose the missing pips are the 432, and on the first round lho plays the 4. The relevant holdings are lho having 43, 42, T2 initially, but from either of the first two he would have played the 4 only half the time, so those two are exactly half as likely as the Tx, and its fifty fifty.

 

In essence, although you are right that restricted choice doesn't apply at trick two, it does apply at trick one, as lho's play is forced from Tx but not from xx, and rho's play of the K is always forced.

 

 

Again, the fool proof way of looking at these is to look at the a-priori holdings. Given that lho has a doublton and rho the K, there are 6 relevant holdings, Tx-Kxx three times, and xx-KTx three times, the card lho plays at trick one does not affect the a priori holdings, so it remains 50-50.

 

 

 

[phil_20686]

Rubens calls the non-K clubs a 'suit within a suit' or a 'subsuit'. There are four non-king clubs, and they will split 2-2 more often than 3-1. We can't cope with a 3-1 remainder either way, so it more or less mirrors a two way finesse.

 

So this type of reasoning is fine, but it is implied that you do not use the a-priori odds for a 2-2/finesse. If all you know is that rho has the K, clearly 3-1 is more likely than 1-3 for example.

 

Again, in this particular case it is easy to work out the relative odds exactly: the drop is exactly as likely to succeed as the finesse.

[JLOGIC]

You are right obv not sure what I was thinking lol.

 

Basically once the king is known it is equally likely that either player has Tx vs xx obviously.

[MrAce]

This logic feels intuitive, but is actually wrong. Suppose the missing pips are the 432, and on the first round lho plays the 4. The relevant holdings are lho having 43, 42, T2 initially, but from either of the first two he would have played the 4 only half the time, so those two are exactly half as likely as the Tx, and its fifty fifty.

 

In essence, although you are right that restricted choice doesn't apply at trick two, it does apply at trick one, as lho's play is forced from Tx but not from xx, and rho's play of the K is always forced.

 

 

Again, the fool proof way of looking at these is to look at the a-priori holdings. Given that lho has a doublton and rho the K, there are 6 relevant holdings, Tx-Kxx three times, and xx-KTx three times, the card lho plays at trick one does not affect the a priori holdings, so it remains 50-50.

 

 

I have to admit i also thought just like Justin, untill i read this reply.

[SteveMoe]

This logic feels intuitive, but is actually wrong. Suppose the missing pips are the 432, and on the first round lho plays the 4. The relevant holdings are lho having 43, 42, T2 initially, but from either of the first two he would have played the 4 only half the time, so those two are exactly half as likely as the Tx, and its fifty fifty.

 

In essence, although you are right that restricted choice doesn't apply at trick two, it does apply at trick one, as lho's play is forced from Tx but not from xx, and rho's play of the K is always forced.

 

 

Again, the fool proof way of looking at these is to look at the a-priori holdings. Given that lho has a doublton and rho the K, there are 6 relevant holdings, Tx-Kxx three times, and xx-KTx three times, the card lho plays at trick one does not affect the a priori holdings, so it remains 50-50.

Echoing others, well said Phil.